Find the least positive integer n for which there exists a set { s_1 ,s_2 ,s_3 ,...,s_n } consisting of n distinct positive integers such that (1−(1/s_1 ))(1−(1/s_2 ))(1−(1/s_3 ))...(1−(1/s


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Question Number 103908 by bobhans last updated on 18/Jul/20
$Find the least positive integer n for which there exists a set { s_1 ,s_2 ,s_3 ,...,s_n } consisting of n distinct positive integers such that (1−(1/s_1 ))(1−(1/s_2 ))(1−(1/s_3 ))...(1−(1/s_n )) = ((51)/(2010)) .$
$F i n d t h e l e a s t p o s i t i v e i n t e g e r n f o r$ $w h i c h t h e r e e x i s t s a s e t {s_{1}, s_{2}, s_{3}, . . ., s_{n}}$ $c o n s i s t i n g o f n d i s t i n c t p o s i t i v e i n t e g e r s$ $s u c h t h a t (1 - \frac{1}{s_{1}}) (1 - \frac{1}{s_{2}}) (1 - \frac{1}{s_{3}}) . . . (1 - \frac{1}{s_{n}})$ $= \frac{51}{2010} .$
Commented by Rasheed.Sindhi last updated on 18/Jul/20

$\frac{51}{2010} = \frac{17}{670}$
Commented by bobhans last updated on 18/Jul/20

$f o r n = ? ? s i r$
	Answered by mr W last updated on 18/Jul/20

	$i g o t :$ $n_{m i n} = 9$ $s_{1} = 67$ $s_{2} = 25$ $s_{3} = 18$ $s_{4} = 11$ $s_{5} = s_{6} = s_{7} = s_{8} = s_{9} = 2$ $b u t t h e y a r e n o t d i s t i n c t!$
	Answered by john santu last updated on 18/Jul/20
	$suppose that for n there exist the desired numbers; we may assumse that s_1 <s_2 <s_3 <...<s_n . surely s_1 >1 since otherwise 1−(1/s_1 ) = 0 . so we have 2≤s_1 ≤s_2 −1≤...≤s_n −(n−1), hence s_i ≥i +1 for each i=1,2,...,n. Therefore ((51)/(2010)) = (1−(1/s_1 ))(1−(1/s_2 ))...(1−(1/s_n )) ≥ (1−(1/2))(1−(1/3))...(1−(1/(n+1)))=(1/2).(2/3).(3/4)...(n/(n+2)) which implies n+1 ≥ ((2010)/(51))=((670)/(17))>39 so n≥ 39. Now we are left to show that n=39 fits. consider the set { 2,3,4,...33,34,...,67 } which contains exactly 39 numbers. we have (1/2).(2/3)...((32)/(33)).((33)/(34))...((39)/(40)).((66)/(67)) = (1/(33)).((34)/(40)).((66)/(67)) = ((17)/(670)) = ((51)/(2010)). hence for n = 39 there exist a desired example . (JS ⊛)$
	$s u p p o s e t h a t f o r n t h e r e e x i s t$ $t h e d e s i r e d n u m b e r s; w e m a y$ $a s s u m s e t h a t s_{1} < s_{2} < s_{3} < . . . < s_{n} .$ $s u r e l y s_{1} > 1 s i n c e o t h e r w i s e$ $1 - \frac{1}{s_{1}} = 0 . s o w e h a v e 2 ⩽ s_{1} ⩽ s_{2} - 1 ⩽ . . . ⩽ s_{n} - (n - 1),$ $h e n c e s_{i} ⩾ i + 1 f o r e a c h$ $i = 1, 2, . . ., n . T h e r e f o r e \frac{51}{2010} =$ $(1 - \frac{1}{s_{1}}) (1 - \frac{1}{s_{2}}) . . . (1 - \frac{1}{s_{n}}) ⩾ (1 - \frac{1}{2}) (1 - \frac{1}{3}) . . . (1 - \frac{1}{n + 1}) = \frac{1}{2} . \frac{2}{3} . \frac{3}{4} . . . \frac{n}{n + 2}$ $w h i c h i m p l i e s n + 1 ⩾ \frac{2010}{51} = \frac{670}{17} > 39$ $s o n ⩾ 39 . N o w w e a r e l e f t t o$ $s h o w t h a t n = 39 f i t s . c o n s i d e r$ $t h e s e t {2, 3, 4, . . . 33, 34, . . ., 67}$ $w h i c h c o n t a i n s e x a c t l y 39$ $n u m b e r s . w e h a v e \frac{1}{2} . \frac{2}{3} . . . \frac{32}{33} . \frac{33}{34} . . . \frac{39}{40} . \frac{66}{67}$ $= \frac{1}{33} . \frac{34}{40} . \frac{66}{67} = \frac{17}{670} = \frac{51}{2010} .$ $h e n c e f o r n = 39 t h e r e e x i s t a$ $d e s i r e d e x a m p l e . (J S ⊛)$
	Commented by bobhans last updated on 18/Jul/20

	$t h a n k y o u$
	Commented by mr W last updated on 18/Jul/20

	$v e r y n i c e!$
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